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Question. Given a convex body $\Omega$, what is the shape of a surface $\Gamma$ of minimal area which divides $\Omega$ into two regions of equal volume?

Background/motivation.

A 2D version of the question was posed by Michael Goldberg in Monthly: find the shortest curve which divides a convex quadrilateral into two equal areas. In the latter case the solution is given by a circular arc perpendicular to two sides of the quadrilateral (or just a segment of a straight line in degenerate situations). This follows from the observation that a circular sector is the shortest curve which, together with two sides of an angle, encloses a fixed area.

Goldberg himself offered a physically intuitive solution to the problem:

The curve may be considered as a restraining member under tension produced by internal fluid pressure in the restricted area. The ends of the curve are free to slide along the sides. Hence, the curve must be normal to two sides of the quadrilateral. Furthermore, since the fluid pressure is uniform, the curve must take the form of a circular arc.

The same approach suggests that in the general multidimensional case the solution is given by a spherical "cap" that intersects $\partial \Omega$ orthogonally. Now, assuming that the intuition is correct, is there a simple formal proof of this conjecture?

Edit. As Sergei Ivanov points out the minimal surface in question need not be a spherical cap for dimensions $>2$.

A modified question: is there always at least one solution to the problem? Should one impose any smoothness conditions on $\partial\Omega$ to guarantee existence?

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  • $\begingroup$ I guess that the condition that "the pressure is equal" was more or less related to saying that $\Gamma\cap B_r(a)$ annihilates the variation of area among perturbations inside $B_r(a)$ with fixed boundary which don't change the volume on the 2 sides of $\Gamma$ in $B_r(a)$. So if the first variation of area was first given by $H=0$, I expect that now it will become $H=const$ as we replace all the possible normal variations with the ones which just have zero mean. More or less we are passing from the main theorem of calculus of variations to the du Bois-Reymond theorem. $\endgroup$
    – Mircea
    Commented Feb 7, 2011 at 15:09
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    $\begingroup$ If you're looking for a way to compute (an approximation) to such a surface in 2 or 3 dimensions, Ken Brakke's freely-available Surface Evolver is the way to go: susqu.edu/brakke/evolver/evolver.html $\endgroup$
    – j.c.
    Commented Feb 7, 2011 at 16:13
  • $\begingroup$ @Mircea and jc: thanks a lot for the helpful comments. $\endgroup$
    – Andrey Rekalo
    Commented Feb 7, 2011 at 17:10

2 Answers 2

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The conjecture as you state it is false. The variational argument (that can be interpreted in terms of fluid pressure if you like) shows that the surface has constant mean curvature and is orthogonal to the boundary. In dimension 2, this means that it is a circular arc, but in higher dimensions there are many more examples of constant mean curvature surfaces.

A small perturbation of the standard ball can produce a convex body such that no sphere is orthogonal to its boundary. For such a body, the surface in question just cannot be a spherical cap.

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  • $\begingroup$ I will try to reformulate the final part of the question. $\endgroup$
    – Andrey Rekalo
    Commented Feb 7, 2011 at 13:34
  • $\begingroup$ Thanks Sergei. Could you explain how this problem can be interpreted in terms of fluid pressure? $\endgroup$
    – A random mathematician
    Commented Nov 21, 2014 at 18:48
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This is a classical problem on which we know the existence, thanks to the Geometric Measure Theory. In space dimension $n$, the solution is a hypersurface which is smooth of constant curvature away from a possible singular set of dimension at most $n-7$. For instance, if $n=3$, it is smooth up to the boundary, and it meets $\partial\Omega$ orthogonally.

If you relax the equal volume condition, and ask for a hypersurface of smaller area that divides $\Omega$ into two pieces of volumes $\epsilon\Omega$ and $(1-\epsilon)\Omega$, then the surface tends to a boundary point $p$ where the curvature has a critical point. See for instance the discussion in the recent PhD Thesis of Paul Laurain (ENS de Lyon).

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  • $\begingroup$ That's amazing! Thank you. Do you think Paul Laurain's thesis is available online? $\endgroup$
    – Andrey Rekalo
    Commented Feb 7, 2011 at 14:39
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    $\begingroup$ Yes it is : umpa.ens-lyon.fr/~plaurain $\endgroup$
    – Denis Serre
    Commented Feb 7, 2011 at 16:00

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